# Determinant of a diagonal block matrix using exterior algebra

Let $$V$$ and $$U$$ be two finite-dimensional vector spaces of dimensions $$n$$ and $$m$$ respectively. For $$f\in \mathrm{End}\,V$$ we define a multilinear, anti-symmetric function: $$V\times ...\times V\to \Lambda^nV,~~(v_1, v_2,\dots, v_n)\mapsto fv_1\wedge fv_2\wedge\dots\wedge fv_n$$ what corresponds to a linear function $$\det f\colon \Lambda^nV\to\Lambda^nV$$.

Now I would like to prove that $$\det f\oplus g = \det f\cdot \det g$$ (i. e. the determinant of a block diagonal matrix is the product of the determinants of blocks).

Naively, if $$v_1, \dots, v_n$$ and $$u_1,\dots, u_m$$ are bases of $$V$$ and $$U$$, we merge them into a basis of $$V\oplus U$$ and: $$(f\oplus g)v_1\wedge \dots (f\oplus g)v_n\wedge (f\oplus g)u_1\wedge (f\oplus g)u_m = \\=(fv_1\wedge \dots\wedge fv_n) \wedge (gu_1\wedge\dots\wedge g u_m) =\\ =\det f\cdot \det g \cdot v_1\wedge\dots\wedge v_n \wedge u_1\wedge\dots\wedge u_m$$ However, I can't bracket out these two terms without some kind of identification between $$\Lambda^{n+m}(V\oplus U)$$ and pressumably $$\Lambda^nV\otimes \Lambda^mU$$. How does one fix this rigorously?

Edit: From Exterior power "commutes" with direct sum we know that there is an isomorphism $$\Lambda ^{n+m}(V\oplus U) \simeq \Lambda^nV\otimes \Lambda^mU$$.

Let $$i_V\colon V\to V\oplus U$$ and $$i_U\colon U \to V\oplus U$$ be the two inclusions. Now if $$v_1, \dots, v_n$$ is a basis of $$V$$ and $$u_1,\dots, u_m$$ is a basis of $$U$$, then: $$i_Vv_1, \dots, i_Vv_n, i_Uu_1, \dots, i_Uu_m$$ is a basis of $$V\oplus U$$. From the definition of $$f\oplus g$$ we know that $$(f\oplus g)i_Vv_i=i_Vfv_i$$ and $$(f\oplus g)i_Uu_i=i_Ugu_i$$. Now define: $$x := (f\oplus g)i_Vv_1\wedge \dots\wedge (f\oplus g)i_Vv_n \wedge(f\oplus g)i_Uu_1\wedge\dots\wedge (f\oplus g)i_Uu_m = \\ =i_Vfv_1\wedge\dots\wedge i_Vfv_n\wedge i_Ugu_1\wedge\dots\wedge i_Ugu_m .$$
We have a linear isomorphism $$\alpha\colon \Lambda^{n+m}(V\oplus U)\to \Lambda^nV\otimes \Lambda^mU$$. Acting with it we get an expression: $$\alpha(x) = (fv_1\wedge\dots\wedge fv_n)\otimes(gu_1\wedge\dots\wedge gu_m) \\ = \det f\cdot \det g\cdot (v_1\wedge\dots \wedge v_n)\otimes (u_1\wedge\dots\wedge u_m)$$ Using its inverse: $$x=(\alpha^{-1}\alpha)(x) = \det f\cdot \det g\cdot i_Vv_1\wedge\dots\wedge i_Vv_n\wedge i_Uu_1\wedge\dots\wedge i_Uu_m ,$$ which proves the claim.
We can reduce clutter resulting from inclusions – the crucial step is to move the calculation from $$\Lambda^{n+m}(V\oplus U)$$ to $$\Lambda^nV\otimes \Lambda^mU$$, and then translate the results back.